Question

In the given figure, line DE || line GF ray EG and ray FG are bisectors of $\angle$DEF and $\angle$DFM respectively. Prove that,
(i) $\angle$DEG = $\frac{1}{2}\angle$EDF (ii) EF = FG.

Answer 1

(i) Given: DE || GF
Now, $\angle$DEF = $\angle$GFM (Corresponding angles as DM is a transversal line)
⇒ 2$\angle$DEG = $\angle$DFG (Ray EG and ray FG are bisectors of $\angle$DEF and $\angle$DFM)
⇒ 2$\angle$DEG = $\angle$EDF (∵ $\angle$EDF = $\angle$DFG, alternate angles as DF is a transversal line)
⇒ $\angle$DEG = $\frac{1}{2}\angle$EDF

(ii) Given: DE || GF
$\angle$DEG = $\angle$EGF (Alternate angles as EG is a transversal line)
∴ $\angle$GEF = $\angle$EGF (∵ $\angle$DEG = $\angle$GEF)
∴ EF = FG (Sides opposite to equal angles)